Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
long long a[maxn][maxn];
long long rminv[maxn], cminv[maxn];
vector<pair<string, long long>> v;
int main(void) {
  cin.tie(0);
  std::ios::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  memset(rminv, 0x3f, sizeof(rminv));
  memset(cminv, 0x3f, sizeof(cminv));
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) cin >> a[i][j];
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) {
      rminv[i] = min(rminv[i], a[i][j]);
    }
  }
  for (long long j = 1; j <= m; j++) {
    for (long long i = 1; i <= n; i++) {
      cminv[j] = min(cminv[j], a[i][j]);
    }
  }
  if (n <= m) {
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= m; j++) a[i][j] -= rminv[i];
      for (long long k = 1; k <= rminv[i]; k++) v.push_back({"row", i});
    }
    for (long long j = 1; j <= m; j++) {
      long long mx = 0x3f3f3f3f;
      for (long long i = 1; i <= n; i++) mx = min(mx, a[i][j]);
      for (long long i = 1; i <= n; i++) a[i][j] -= mx;
      for (long long k = 1; k <= mx; k++) v.push_back({"col", j});
    }
    bool flag = 1;
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= m; j++) {
        if (a[i][j] != 0) {
          flag = 0;
        }
      }
    }
    if (0 == flag) {
      cout << "-1"
           << "\n";
    } else {
      cout << v.size() << "\n";
      for (auto i : v) {
        cout << i.first << " " << i.second << "\n";
      }
    }
  } else {
    for (long long j = 1; j <= m; j++) {
      for (long long i = 1; i <= n; i++) a[i][j] -= cminv[j];
      for (long long k = 1; k <= cminv[j]; k++) v.push_back({"col", j});
    }
    for (long long i = 1; i <= n; i++) {
      long long mx = 0x3f3f3f3f;
      for (long long j = 1; j <= m; j++) mx = min(mx, a[i][j]);
      for (long long j = 1; j <= m; j++) a[i][j] -= mx;
      for (long long k = 1; k <= mx; k++) v.push_back({"row", i});
    }
    bool flag = 1;
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= m; j++) {
        if (a[i][j] != 0) {
          flag = 0;
        }
      }
    }
    if (0 == flag) {
      cout << "-1"
           << "\n";
    } else {
      cout << v.size() << "\n";
      for (auto i : v) {
        cout << i.first << " " << i.second << "\n";
      }
    }
  }
  return 0;
}